make a 99% confidence interval for the population mean, μ? thanks?
A random sample of 25 life insurance policyholders showed that the average premium they pay on their life insurance policies is $504 per year with a standard deviation s of $78.
Assuming that the life insurance policy premiums for all life insurance policyholders have a normal distribution, make a 99% confidence interval for the population mean, μ (Round interval bounds to 2 decimal places).
More Questions From This User See All
Answers & Comments
Verified answer
P (460.3677 < μ < 547.6323) = P (-2.7969 < t < 2.7969) = 99.0000% C.I. d.f.=24
by capacity of certainty the development length is small a t-statistic must be used to calculate the self belief era. the final expression is: Xbar ± t?/2[s/?n] For ninety 5% self belief t = 2.262 for ?/2 and 9 levels of freedom. X bar = 0.32, and s = 0.09 So the self belief era is: 0.32 ± 2.262[0.09/3.sixteen] = 0.256 and 0.384 particular, it must be easily finding for the producer to end that the inhabitants advise tread last replaced into 0.30 inches for the reason that the cost is included interior the self belief era.